In the following figure AB - AC and AD is perpendicular to BC - BE bisects angle B and EF is perpendicular to AB Prove thatED - EF

Question

Toppr Admin · Accepted Answer

In - -Delta EFB- and -Delta EDB - -angle EFB - -angle EDB -160- -160- -160- -160- -160- - both are - 90 -o-EB - EB-160- -160- -160- -160- -160- -160- -160- -160- -160-160- - common - -angle FBD - -angle DBE -160- -160- -160- -160- -160- -160- -160- -given - -Delta EFB -cong -Delta EDB -160- -160- -160-160- By angle-side-angle congruency Also -ED - EF-160-160-160- -By C-P-C-T-

In the following figure AB = AC and AD is perpendicular to BC . BE bisects angle B and EF is perpendicular to AB
Prove that
ED = EF

In $$ \Delta EFB$$ and $$\Delta EDB $$
$$ \angle EFB = \angle EDB $$ ( both are $$ 90 ^{o}$$)
$$EB = EB$$ [ common ]
$$ \angle FBD = \angle DBE $$ (given )

$$ \Delta EFB \cong \Delta EDB $$ By angle-side-angle congruency

Also $$ED = EF$$ (By C.P.C.T)

In the following figure AB = AC and AD is perpendicular to BC . BE bisects angle B and EF is perpendicular to AB Prove thatED = EF

In $$ \Delta EFB$$ and $$\Delta EDB $$$$ \angle EFB = \angle EDB $$ ( both are $$ 90 ^{o}$$)$$EB = EB$$ [ common ]$$ \angle FBD = \angle DBE $$ (given )$$ \Delta EFB \cong \Delta EDB $$ By angle-side-angle congruency Also $$ED = EF$$ (By C.P.C.T)

In the following figure AB = AC and AD is perpendicular to BC . BE bisects angle B and EF is perpendicular to AB
Prove that
ED = EF

In $$ \Delta EFB$$ and $$\Delta EDB $$
$$ \angle EFB = \angle EDB $$ ( both are $$ 90 ^{o}$$)
$$EB = EB$$ [ common ]
$$ \angle FBD = \angle DBE $$ (given )

$$ \Delta EFB \cong \Delta EDB $$ By angle-side-angle congruency

Also $$ED = EF$$ (By C.P.C.T)