Question

given- CF perpendicular to AB and BE is perpendicular to AC

Prove AD is perpendicular to BC

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Answer 1

Step-by-step explanation:

i think u have drawn a circle around the

so it is circumcircle

in bfc and ecb

<bfc = <bec

bc is a common side

theorem:in any triangle all the three perpendiculars are concurrent

hence it is given that cf _|_ ab and be _|_ ac

so the third perpendicular will also pass through that point

point of concurrence of perpendiculars is called circumcentre

let circumcentre be 0

oc=ob (circumradius)

therefore bfc is congruent to ecb

by cpct <adb=<adc

<adb+<adc=180°

as both r equal both r 90°