Question 1: In the adjoining figure, and are two points on equal sides and of such that . Prove that .
Answer:
Consider and
(given)
is common
(given)
Therefore (by S.A.S theorem)
Hence
Question 2: If is the mid point of the hypotenuse of a right triangle , prove that .
Answer:
Construct ABD such that ABD = DE
Consider and
(given)
(vertically opposite angles)
(constructed)
Therefore (by S.A.S theorem)
and
(interior angles on the same side of the transversal are supplementary)
Therefore and are right angle triangles.
Therefore
is common
Therefore
. Hence proved.
Question 3: In a quadrilateral and bisects . Show that . What can you say about and .
Answer:
Consider and
(given)
(given)
is common
Therefore (by S.A.S theorem)
Question 4: Prove that is isosceles if any of the following holds: (i) Altitude bisects (ii) Median is perpendicular to the base .
Answer:
(i) If
Consider and
is common
(altitude is perpendicular to the base)
(given)
Therefore
Therefore is isosceles triangle.
(ii) If Median is perpendicular to the base
is common
(since is the median, is the midpoint of )
Therefore
Therefore is isosceles triangle.
Question 5: In the adjoining figure and . Prove that , and .
Answer:
Consider and
is given
given
(is common)
Therefore
Therefore and
Question 6: In the adjoining figure, , and . Prove that
Answer:
Given , and
Therefore (by S.A.S theorem)
Question 7: In a , and and are midpoints of the sides and respectively. Prove that .
Answer:
Given
Consider and
Therefore
Question 8: In the adjoining figure, is a square and is an equilateral triangle. Prove that (i) (ii)
Answer:
Consider and
(equilateral triangle)
(sides of a square)
Since and
Therefore
In
Therefore
Question 9: Prove that the median of an equilateral triangle are equal.
Answer:
Given:
is equilateral
Therefore
To prove:
Consider and
is an equilateral triangle)
is an equilateral triangle)
(medians bisect the opposite side. Since is an equilateral triangle the sides are equal)
Therefore
Now consider and
is an equilateral triangle)
is an equilateral triangle)
Therefore
Hence
Hence we can say that
Question 10: is a line segment. and are points on opposite sides of such that each of them is equidistant from the points and . Show that the line is perpendicular bisector of .
Answer:
Consider and
(given)
(bisector)
is common
Therefore
Therefore
Similarly in and
is common
Therefore
Therefore is a perpendicular bisector of
Question 11: In the adjoining figure, and , find the ratio
Answer:
Since
Also since
Now
Question 12: In the adjoining figure, and . Prove that and .
Answer:
Given:
Since
Therefore
Therefore in and , we have
is common
Therefore (A.S.A. theorem)
Hence and
Question 13: In the adjoining figure, and . Prove that and hence and $latex $latex .
Answer:
Given:
Consider and
(Proved above)
(given)
(given)
Therefore (A.S.A theorem)
Therefore and
Question 14: In the adjoining figure, it is given that and . Prove that .
Answer:
Consider , we have
… … … … … (i)
We have (vertically opposite angles)
… … … … … (ii)
Subtracting (ii) from (i) we get
Therefore in and , we have
(common angle)
(given) and
Therefore (A.S.A. theorem)
Question 15: and are bisectors of and of an isosceles with . Prove that .
Answer:
Given: bisects the
Similarly, bisects the
Since is an isosceles triangle,
Consider and
is common
Since
Hence (A.S.A theorem)