Question 1: In the adjoining figure, $\triangle ABC$ is an obtuse triangle, obtuse at $\angle B$. If $AD \perp CB$, prove that $AC^2 = AB^2 + BC^2 + 2 BC \times BD$.

Given $\triangle ABC$ is an obtuse triangle, $\angle B$ is obtuse angle

In $\triangle ADB$, we have

$AB^2 = AD^2 + DB^2$         (by Pythagoras theorem)

In $\triangle ADC$, we have

$AC^2 = AD ^2 + DC^2$

$\Rightarrow AC^2 = AD^2 + (DB + BC)^2$

$\Rightarrow AC^2 = (AD^2 + DB^2) + BC^2 + 2 DB \times BC$

$\Rightarrow AC^2 = AB^2 + BC^2 + 2 DB \times BC$

Hence proved.

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Question 2: In the adjoining figure, $\angle B$ of $\triangle ABC$ is acute angle and $AD \perp BC$. Prove that $AC^2 = AB^2 + BC^2 - 2 BC \times BD$

In $\triangle ADB, \angle D = 90^o$

Therefore $AB^2 = AD^2 + BD^2$         (by Pythagoras theorem)

Similarly, in $\triangle ADC$, we have

$AC^2 = AD^2 + DC^2$         (by Pythagoras theorem)

$\Rightarrow AC^2 = AD^2 + (BC-BD)^2$

$\Rightarrow AC^2 = AD^2 + BC^2 + BD^2 - 2 BC \times BD$

$\Rightarrow AC^2 = AB^2 + BC^2 - 2 BC \times BD$

Hence proved.

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Question 3: Prove in any triangle, the sum of the square of any two sides is equal to twice the square of half of the third side together with twice the square of the median which bisects the third side.

Construction: Altitude, $\angle AED = 90^o$

The $\triangle ADB$, is obtuse angle triangle, and $\triangle ADC$ is an acute angle triangle.

In $\triangle ABD, \angle D$ is obtuse and $AE \perp BD$.

Therefore $AB^2 = AD^2 + BD^2 + 2 BD \times DE$ … … … … … i)

Similarly, in $\triangle ADC$ is acute at $\angle D$ and $AE \perp CD$

Therefore $AC^2 = AD^2 + DC^2 - 2 DC \times DE$

$\Rightarrow AC^2 = AD^2 + BD^2 - 2 BD \times DE$ (since $CD = BD$) … … … … … ii)

Adding i) and ii) we get

$AB^2 + AC^2 = 2 AD^2 + 2 BD^2$

$\Rightarrow AB^2 + AC^2 = 2 (AD^2 + BD^2)$ … … … … … ii)

Putting $BD = \frac{1}{2} BC$   in iii) we get

$AB^2 + AC^2 = 2 AD^2 + 2 (\frac{1}{2} BC^2)$

Hence proved.

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Question 4: Prove that three times the sum of the square of the sides of a triangle is equal to 4 times the sum of square of medians of the triangle.

Given $\triangle ABC$ and $AD, BE$ and $CF$ are medians as shown in the diagram.

In $\triangle ABC$, with $AD$ as the median of $BC$, we have

$AB^2 + AC^2 = 2 (AD^2 + BD^2)$

$\Rightarrow AB^2 + AC^2 = 2 \{AD^2 + (\frac{BC}{2})^2 \}$

$\Rightarrow 2(AB^2 + AC^2 ) = 4 AD^2 + BC^2$ … … … … … i)

Similarly with $BE$ bisecting $AC$ and $CF$ bisecting $AB$, we get

$2 (AB^2 + BC^2) = 4 BE^2 + AC^2$

and $2 (AC^2 + BC^2) = 4CF^2 + AB^2$

Adding i), ii) and iii) we get

$4(AB^2 + BC^2 + CA^2) = 4(AD^2 + BE^2 + CF^2) + BC^2 + CA^2 + AB^2$

$\Rightarrow 3 (AB^2 + BC^2 + CA^2) = 4 (AD^2 + BE^2 + CF^2)$

Hence proved.

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Question 5: $P$ and $Q$ are mid points of side $CA$ and $CB$ respectively in $\triangle ABC$. $\angle C$ is $90^o$. Prove that (i) $4 AQ^2 = 4 AC^2 + BC^2$ (ii) $4 BP^2 = 4 BC^2 + AC^2$ (iii) $4 (AQ^2 + BP^2) = 5 AB^2$

(i) In $\triangle AQC, \angle C = 90^o$

$AQ^2 = AC^2 + QC^2$ (by Pythagoras theorem)

$\Rightarrow 4 AQ^2 = 4 AC^2 + 4 QC^2$ (multiply by 4 on both sides)

$\Rightarrow 4AQ^2 = 4 AC^2 + (2BC^2)$ (since $BC = 2 QC$)

(ii) In $\triangle BPC$, $\angle C= 90^o$

$BP^2 = BC^2 + CP^2$

$\Rightarrow 4 BP^2 = 4 BC^2 + 4 CP^2$

$\Rightarrow 4 BP^2 = 4 BC^2 + AC^2$ (since $AC = 2CP$)

(iii) From i) and ii) we have

$4 AQ^2 = 4 AC^2 + BC^2$

$4 BP^2 = 4 BC^2 + AC^2$

$4 AQ^2 + 4 BP^2 = (4 AC^2 + BC^2) + (4 BC^2 + AC^2)$

$\Rightarrow 4 (AQ^2 + BP^2) = 5 (AC^2 + BC^2)$ (since we have $AB^2 = AC^2 + BC^2$)

$\Rightarrow 4(AQ^2 + BP^2) = 5 AB^2$

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Question 6: In the adjoining figure, $\triangle ABC$ is right angles at $\angle B$. $AD$ and $CE$ are two medians drawn from vertices $A$ and $C$ respectively. If $AC$ is $5 \ cm$ and $AD = \frac{3\sqrt{5}}{2} \ cm$, find the length of $CE$.

In $\displaystyle \triangle ABD$

$\displaystyle AD^2 = AB^2 + BD^2$

$\displaystyle \Rightarrow AD^2 = AB^2 + (\frac{BC}{2})^2$

$\displaystyle \Rightarrow AD^2 = AB^2 + \frac{1}{4} BC^2$

Similarly in $\displaystyle \triangle BCE$

$\displaystyle CE^2 = BC^2 + \frac{1}{4} AB^2$

$\displaystyle \Rightarrow AD^2 + CE^2 = AB^2 + \frac{1}{4} BC^2 + BC^2 + \frac{1}{4} AB^2$

$\displaystyle \Rightarrow AD^2 + CE^2 = \frac{5}{4} (AB^2 + BC^2)$

$\displaystyle \Rightarrow AD^2 + CE^2 = \frac{5}{4} AC^2$

$\displaystyle \Rightarrow ( \frac{3 \sqrt{5}}{2} )^2 + CE^2 = \frac{5}{4} (5)^2$

$\displaystyle \Rightarrow CE^2 = \frac{125}{4} - \frac{45}{4} = 20$

$\displaystyle \Rightarrow CE = 2\sqrt{5} cm$

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Question 7: In $\triangle ABC, \angle C = 90^o$. Let $BC = a, CA = b$ and $AB = c$ and let $p$ be the length of the perpendicular from $C$ to $AB$. Prove that

(i) $cp = ab$

(ii) $\frac{1}{p^2}$ $=$ $\frac{1}{a^2}$ $+$ $\frac{1}{b^2}$

Consider the figure

(i) $\displaystyle CD \perp AB$, then $CD = p$

$\displaystyle \text{Area of } \triangle ABC = \frac{1}{2} \times AB \times CD = \frac{1}{2} cp$

$\displaystyle \text{Also the area of } \triangle ABC = \frac{1}{2} \times BC \times AC = \frac{1}{2} ab$

$\displaystyle \text{Therefore } \frac{1}{2} cp = \frac{1}{2} ab$

$\displaystyle \Rightarrow cp = ab$

(ii) In $\displaystyle \triangle ABC, \angle C = 90^o$, therefore

$\displaystyle AB^2 = BC^2 + AC^2$

$\displaystyle \Rightarrow c^2 = a^2 +b^2$

$\displaystyle \text{(since } cp = ab \Rightarrow c = \frac{ap}{p} )$

$\displaystyle \Rightarrow ( \frac{ab}{p} )^2 = a^2 + b^2$

$\displaystyle \Rightarrow \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2}$

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Question 8: In an equilateral triangle with side $a$, prove that (i) Altitude $=$ $\frac{a\sqrt{3}}{2}$   (ii) Area $=$ $\frac{\sqrt{3}}{4}$ $a^2$

(i) Draw $\displaystyle AD \perp BC$

$\displaystyle D$ is the mid point since $\displaystyle \triangle ABC$ is equilateral with $\displaystyle AB = AC = BC$

$\displaystyle \Rightarrow AB = a, BD = \frac{1}{2} BC = \frac{a}{2}$

Since $\displaystyle \triangle ABD$ is right angled at $\displaystyle \angle D$

$\displaystyle \therefore AB^2 = AD^2 + BD^2$

$\displaystyle \Rightarrow a^2 = AD^2 + ( \frac{a}{2} )^2$

$\displaystyle \Rightarrow AD^2 =a^2 - ( \frac{a}{2} )^2$

$\displaystyle \Rightarrow AD^2 = a^2 - \frac{a^2}{4} = \frac{3}{4} a^2$

$\displaystyle \Rightarrow AD = \frac{\sqrt{3}}{2} a$

(ii) Now Area of $\displaystyle \triangle ABC = \frac{1}{2} \times BC \times AD = \frac{1}{2} \times a \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4} a^2$

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Question 9: $ABC$ is a triangle in which $AB = AC$ and $D$  is any point on $BC$. Prove that $AB^2 - AD^2 = BD.CD$

Draw $AE \perp BC$

In $\triangle AEB$ and $\triangle AEC$, we have

$AB = AC$

$AE$ is common

and $\angle B = \angle C$ (since $AB = AC$)

$\therefore \triangle AEB \cong \triangle AEC$

$\Rightarrow BE = CE$

Since $\triangle AED$ and $\triangle ABE$ are right angles at $E$

$\therefore AD^2 = AE^2 + DE^2$

$AB^2 = AE^2 + BE^2$

$\Rightarrow AB^2 - AD^2 = BE^2 - DE^2$

$\Rightarrow AB^2 - AD^2 = (BE - DE)(BE + DE)$

$\Rightarrow AB^2 - AD^2 = (BE - DE)(CE + DE)$

$\Rightarrow AB^2 - AD^2 = BD .CD$

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Question 10: From a point $O$ in the interior of $\triangle ABC$, perpendiculars $OD, OE$ and $OF$ are drawn to side $BC, CA$ and $AB$ respectively. Prove:

i) $AF^2 + BD^2 + CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2$

ii) $AF^2 + BD^2 + CE^2 = AE^2 + CD^2 + BF^2$

Let $O$ be the point inside $\triangle ABC$ and let $OD \perp BC, OE \perp CA$ and $OF \perp AB$

i) In $\triangle OFA, \triangle ODB$ and $\triangle OEC$ we have

$OA^2 = AF^2 + OF^2$

$OB^2 = BD^2 + OD^2$

and $OC^2 = CE^2 + OE^2$

$OA^2 + OB^2 + OC^2 = AF^2 + OF^2 + BD^2 + OD^2 + CE^2 + OE^2$

$AF^2 +BD^2 +CE^2 = OA^2 + OB^2 + OC^2 - OD^2 - OE^2 - OF^2$

ii) In right triangle $\triangle ODB$ and $\triangle ODC$ we have

$OB^2 = OD^2 + BD^2$

$OC^2 = OD^2 + CD^2$

$OB^2 - OC^2 = (OD^2 + BD^2)-(OD^2 + CD^2)$

$\Rightarrow OB^2 - OC^2 = BD^2 - CD^2$

Similarly, $OC^2 - OA^2 = CE^2 - AE^2$

$OA^2 - OB^2 = AF^2 - BF^2$

Adding i) , ii) and iii)

$(OB^2 - OC^2) + (OC^2 - OA^2) + (OA^2 - OB^2) = (BD^2 - CD^2) + (CE^2 - AE^2) + (AF^2 - BF^2)$

$\Rightarrow (BD^2 + CE^2 + AF^2) - (AE^2 + CD^2 + BF^2) = 0$

$\Rightarrow AF^2 + BD^2 + CE^2 = AE^2 + BF^2 + CD^2$

Hence proved.

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Question 11: A point $O$ inside a rectangle $ABCD$ is joined with each of the vertices $A, B, C$ and $D$. Prove that $OB^2 + OD^2 = OC^2 + OA^2$

Join $O$ to $A, B, C$ and $D$

Draw a line $EF \parallel AB$ through $O$

In $\triangle OEA$ and $\triangle OFC$ we have

$OA^2 = OE^2 + AE^2$

$OC^2 = OF^2 + CF^2$

$\Rightarrow OA^2 + OC^2 = OE^2 + AE^2 + OF^2 + CF^2$

$\Rightarrow OA^2 + OC^2 = OE^2 + OF^2 + AE^2 + CF^2$ … … … … … i)

Similarly, in $\triangle OFB$ and $\triangle OFC$ we have

$OB^2 = OF^2 + FB^2$

$OD^2 = OE^2 + DE^2$

$\Rightarrow OB^2 + OD^2 = (OF^2 + FB^2) + (OE^2 + DE^2)$

$\Rightarrow OB^2 + OD^2 = OE^2 + OF^2 + DE^2 + BF^2$

$\Rightarrow OB^2 + OD^2 = OE^2 + OF^2 + CF^2 + AE^2$ … … … … … ii)

Since $DE = CF$ and $AE = BF$

Therefore from i) and ii) we get

$OA^2 + OC^2 = OB^2 + OD^2$

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Question 12: In $\displaystyle \triangle ABC, AC > AB, D$ is mid point of $\displaystyle BC$ and $\displaystyle AE \perp BC$. Prove:

i) $\displaystyle AC^2 = AD^2 + BC.DE + \frac{1}{2} BC^2$

ii) $\displaystyle AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2$

iii) $\displaystyle AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2$

Given $\displaystyle \angle AED = 90^o$

$\displaystyle \therefore \angle ADE < 90^o$ and $\displaystyle \angle ADC > 90^o$

i) In $\displaystyle \triangle ADC, \angle ADC$ is an obtuse angle

$\displaystyle \therefore AC^2 = AD^2 + DC^2 + 2 DC.DE$

$\displaystyle \Rightarrow AC^2 = AD^2 + ( \frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC .DE$

$\displaystyle \Rightarrow AC^2 = AD^2 + \frac{1}{4} BC^2 + BC.DE$

$\displaystyle \Rightarrow AC^2 =AD^2 + BC.DE + \frac{1}{4} BC^2$

ii) In $\displaystyle \triangle ABD, \angle ADE$ is an acute angle

$\displaystyle \therefore AB^2 = AD^2 + BD^2 - 2 BD .DE$

$\displaystyle \Rightarrow AB^2 = AD^2 + (\frac{1}{2} BC)^2 + 2 . \frac{1}{2} BC.DE$

$\displaystyle \Rightarrow AB^2 = AD^2 + \frac{1}{4} BC^2 - BC .DE$

$\displaystyle \Rightarrow AB^2 = AD^2 - BC.DE + \frac{1}{4} BC^2$

iii) From i) and ii) we get

$\displaystyle AB^2 + AC^2 = 2 AD^2 + \frac{1}{2} BC^2$

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Question 13: In an equilateral $\triangle ABC, BC$ is trisected at $D$. Prove that $9 AD^2 = 7 AB^2$

$\displaystyle BD = \frac{1}{3} BC$

Draw $\displaystyle AE \perp BC$

Join $\displaystyle AD$

In $\displaystyle \triangle AEB$ and $\displaystyle \triangle AEC$, we have $\displaystyle AB = AC$

$\displaystyle \angle AEB = \angle AEC = 90^o$

and $\displaystyle AE$ is common

$\displaystyle \therefore \triangle AEB \sim \triangle AEC$

Therefore we have

$\displaystyle BD = \frac{1}{2} BC$

$\displaystyle DC = \frac{2}{3} BC$

$\displaystyle BE = EC = \frac{1}{2} BC$

Since $\displaystyle \angle C = 60^o$ (equilateral triangle)

Therefore $\displaystyle \triangle ADC$ is an acute triangle

$\displaystyle \therefore AD^2 = AC^2 + DC^2 - 2 DC \times EC$

$\displaystyle \Rightarrow AD^2 = AC^2 + ( \frac{2}{3} BC)^2 - 2 . \frac{2}{3} BC \times \frac{1}{2} BC$

$\displaystyle \Rightarrow AD^2 = AC^2 + \frac{4}{9} BC^2 - \frac{2}{3} BC^2$ (since $\displaystyle AB = AC = BC$)

$\displaystyle \Rightarrow AD^2 = AB^2 + \frac{4}{9} AB^2 - \frac{2}{3} AB^2$

$\displaystyle \Rightarrow AD^2 = \frac{7}{9} AB^2$

$\displaystyle \Rightarrow 9AD^2 = 7 AB^2$

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Question 14: In adjoining figure $\triangle ABC$ is a right angled triangle.  $\angle B = 90^o$ and point $D$ and $E$ trisect $BC$. Prove  $8AE^2 = 3AC^2 + 5AD^2$

Given $BD = DE = CE$

Let $BD = DE = CE = x$

$\Rightarrow BE = 2x$ and $BC = 3x$

In $\triangle ABD, \triangle ABE$ and $\triangle ABC$ we have

$AD^2 = AB^2 + BD^2$

$\Rightarrow AD^2 = AB^2 + x^2$

$AE^2 = AB^2 + BE^2$

$\Rightarrow AE^2 = AB^2 + 4x^2$

and $AC^2 = AB^2 + BC^2$

$\Rightarrow AC^2 = AB^2 + 9x^2$

Now $8AE^2 - 3 AC^2 - 5 AD^2 = 8(AB^2 + 4x^2) - 3(AB^2 + 9x^2) - 5(AB^2 + x^2)$

$\Rightarrow 8AE^2 - 3 AC^2 - 5 AD^2 = 0$

$\Rightarrow 8AE^2 = 3 AC^2 + 5AD^2$

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Question 15: $ABC$ is a right angled triangle. $\angle A = 90^o$. A circle is inscribed in it. The lengths of the two sides containing the right angle are $6 \ cm$and $8 \ cm$. Find the radius of the circle.

In $\triangle ABC$, we have

$BC^2 = AB^2 + AC^2$

$BC^2 = 6^2 + 8^2 = 100$

$\Rightarrow BC = 10 \ cm$

Now, Area of $\triangle ABC = Area \ of \ \triangle OAB + Area \ of \ \triangle OBC + Area \ of \ \triangle OCA$

$\displaystyle \Rightarrow \frac{1}{2} AB \times AC = \frac{1}{2} AB \times r + \frac{1}{2} BC \times r + \frac{1}{2} CA \times r$

$\Rightarrow 48 = 6r + 10r+8r$

$\Rightarrow r = 48 = 24 r$

$\Rightarrow r = 2 \ cm$

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Question 16: In adjoining figure, $\displaystyle D$ is the mid point of $\displaystyle BC$ and $\displaystyle AE \perp BC$. If $\displaystyle BC = a, AC = b, AB = c, ED = x, AD = p$ and $\displaystyle AE = h$, prove that:

i) $\displaystyle b^2 = p^2 + ax + \frac{a^2}{4}$

ii) $\displaystyle c^2 = p^2 - ax + \frac{a^2}{4}$

iii) $\displaystyle b^2 + c^2 = 2p^2 + \frac{a^2}{2}$

$\displaystyle \text{In } \triangle AEB: h^2 + ( \frac{a}{2} -x)^2 = c^2$ … … … … … … i)

$\displaystyle \text{In } \triangle AED: h^2 +x^2 = p^2$ … … … … … … ii)

$\displaystyle \text{In } \triangle AEC: h^2 + (x + \frac{a}{2} )^2 = b^2$ … … … … … … iii)

(ii) From i) and ii)

$\displaystyle p^2 -x^2 + ( \frac{a}{2} -x)^2 = c^2$

$\displaystyle p^2 -x^2 + \frac{a^2}{4} + x^2 - ax = c^2$

$\displaystyle p^2 + \frac{a^2}{4} - ax = c^2$

or $\displaystyle c^2 = p^2 - ax + \frac{a^2}{4}$

(i) from ii) and iii) we get

$\displaystyle p^2 -x^2 + (x + \frac{a}{2} )^2 = b^2$

$\displaystyle p^2 -x^2 + x^2 + \frac{a^2}{4} + ax = b^2$

$\displaystyle p^2 + \frac{a^2}{4} + ax = b^2$

(iii) Adding (ii) and (iii) we get

$\displaystyle b^2 + c^2 = p^2 + ax + \frac{a^2}{4} + p^2 -ax + \frac{a^2}{4}$

$\displaystyle b^2 + c^2 = 2p^2 + \frac{a^2}{2}$

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Question 17: In the adjoining figure, $\angle B < 90^o, AD \perp BC$. Prove:

i) $b^2 = h^2 + a^2 + x^2 -2ax$

ii) $b^2 = a^2 + c^2 -2ax$

In $\triangle ADB: h^2 + x^2 = c^2$ … … … … … … i)

In $\triangle ADC: h^2 + (a-x)^2 = b^2$ … … … … … … ii)

(i) We know, $AC^2 = AB^2 + BC^2 - 2 BC \times BD$

$b^2 = c^2 +a^2 -2.a.x$

$b^2 = h^2 + x^2 + a^2 -2ax$

or $b^2 = h^2 + a^2 + x^2 - 2ax$

(ii) From  i and ii)

$c^2 - x^2 + (a-x)^2 = b^2$

$c^2 - x^2 + a^2 + x^2 - 2ax = b^2$

$c^2 + a^2 -2ax = b^2$

or $b^2 = a^2 + c^2 -2ax$

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Question 18: $\triangle ABD$ is right angled. $\angle A = 90^o$ and $AC \perp BD$. Prove that:

i) $AB^2 = BC.DC$

ii) $AC^2 = BC.DC$

iii) $AD^2 = BD.CD$

iv) $\displaystyle \frac{AB^2}{AC^2} + \frac{BD}{DC}$

In $\triangle ABC: AB^2 = AC^2 + BC^2$ … … … … … … i)

In $\triangle ACD: AD^2 = AC^2 + DC^2$ … … … … … … ii)

In $\triangle ABD: DB^2 = AB^2 + AD^2$ … … … … … … iii)

(ii) $AB^2 + AD^2 = 2 AC^2 + BC^2 + DC^2$

$DB^2 = 2 AC^2 + BC^2 + DC^2$

$(DC + BC)^2 = 2 AC^2 + BC^2 + DC^2$

$DC^2 + BC^2 + 2 DC.BC = 2 AC^2 + BC^2 + DC^2$

$AC^2 = DC.BC$

(i) $AB^2 = DC.BC + BC^2$

$AB^2 = BC (DC + BC)$

$AB^2 = BC.DB$

(iii) $AD^2 = DC.BC + DC^2$

$AD^2 = DC (BC + DC)$

$AD^2 = DC.DB$

(iv) Dividing i) by ii) we get

$\displaystyle \frac{AB^2}{AC^2} + \frac{BC.DB}{DC.BC} + \frac{BD}{DC}$

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