Question

In the adjacent figure ∆ABC, D is the midpoint of BC. DE perpendicular to AB, DF perpendicular to AC and DE=DF. show that ∆BED is congruence to ∆CFD
​ pls answer fast

Answer 1

in ∆CFD and ∆BFD

ED=DF (GIVEN)

BD=CD (GIVEN)

Angle E = angle F =90° (GIVEN)

hence ∆CFD and ∆BFD are congruence (by SAS)

Answer 2

Answer:

△CFD

Step-by-step explanation:

From △BED and △CFD

BD=CD (D is a midpoit of BC)

DE=DF (Given)

∠DEB=∠DFC (DE⊥AB and DF⊥AC)

Thus,

△BED≅△CFD (By RHS)

Hopefully it helps you, ARMY.

Oops, sorry. I forgot to tell you,

Good Morning, ARMY.

Purple You ARMY.