Mathematics
Answer
Given,
⇒ BC = DE
Adding CD on both sides, we get :
⇒ BC + CD = DE + CD
⇒ BD = CE.
In △ ABD and △ FEC,
⇒ ∠ABD = ∠FEC (Both equal to 90°)
⇒ AB = EF (Given)
⇒ BD = CE (Proved above)
∴ ∆ ABD ≅ ∆ FEC (By S.A.S. axiom)
We know that,
Corresponding parts of congruent triangles are equal.
∴ AD = FC.
Hence, proved that AD = FC.
Related Questions
AD and BC are equal perpendiculars to a line segment AB. If AD and BC are on different sides of AB prove that CD bisects AB.
In △ ABC, AB = AC and the bisectors of angles B and C intersect at point O. Prove that :
(i) BO = CO
(ii) AO bisects angle BAC.
A point O is taken inside a rhombus ABCD such that its distances from the vertices B and D are equal. Show that AOC is a straight line.
In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ.
Prove that △ APX ≅ △ CQY.
![In the given figure, ABCD is a rectangle and X and Y are the mid-points of the sides DC and AB respectively. P and Q are the points of AD and BC respectively such that DP = BQ. Triangles [Congruency in Triangles], Concise Mathematics Solutions ICSE Class 9.](https://cdn1.knowledgeboat.com/img/cm9/q12-test-yourself-c9-icse-class-9-concise-maths-upd-2027-new-updated-658x429.png)