Question

EFG is a right triangle with EF = EG. The bisector of angle E meets FG at H.Prove that FG = 2 EH.

Answer 1

Given :- EFG is a right triangle with EF = EG. The bisector of angle E meets FG at H. Prove that FG = 2 EH. ?

Solution :-

in Right angled ∆FEG, we have,

→ EF = EG (given)

→ ∠FEG = 90° (given)

so,

→ ∠EFG = ∠FGE = 45° (angle opposite to equal sides are equal.)

now, given that, angle bisector of ∠E meets FG at H .

we know that, angle bisector divides ∠E in two equal parts .

then,

→ ∠FEH = ∠HEG = 45°

now, in ∆FEH we have,

→ ∠FEH = 45°

→ ∠HFE = 45°

then,

• side opposite to equal angles are equal .

therefore,

→ FH = EH ----------- (1)

similarly, in ∆HEG we have,

→ ∠HEG = 45°

→ ∠HGE = 45°

then,

• side opposite to equal angles are equal .

therefore,

→ HG = EH . ------------ (2)

adding (1) and (2) we get,

→ FH + HG = EH + EH

→ FG = 2EH (Proved.)

Learn more :-

In ABC, AD is angle bisector,

angle BAC = 111 and AB+BD=AC find the value of angle ACB=?

https://brainly.in/question/16655884