Question

In the adjacent figure A ABC
D is the midpoint of Be
PE LAB, DELAC and DE=DF
Show the A BED = A CFD​

Answer 1

Answer:

Given data:

In ∆ABC,

D is the midpoint of BC

∴ BD = DC ….. (i)

DE ⊥ AB ….. (ii)

DF ⊥ AC ….. (iii)

DE = DF ….. (iv)

To show: ∆BED is congruent to ∆CFD

In ∆BED and ∆CFD, we have

BD = DC [from (i)]

∠BED = ∠CFD = 90° [from (ii) & (iii)]

DE = DF [from (iv)]

Thus, by RHS congruence i.e., Right-angle-Hypotenuse-Side congruence in which, if two right-angled triangles have their hypotenuses and a pair of shorter sides are equal in length then the triangles congruent, we can say

∆BED ≅ ∆CFD

Hence proved  

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