answer the 4th question all part
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a) Ag ADB=ag EDC (vert.opp angle)
ag CED=ag DBA (alternate interior angle)
by AA similarity rule tg. ADB=tg. CDE
b) when the anger ECD= angle DAB (interior angle)
angle BAD=angle ABD (opposite sides of triangle is equal)
In tg. BDC
angle DBC=angle DCB (opposite sides of triangle is equal)
so, tg. ABC =tg. ECB
c) we prove angle ABC = angle ECB is equal
when angle ABC is 90° so angle ECB is 90°
pls mark as brainly
ag CED=ag DBA (alternate interior angle)
by AA similarity rule tg. ADB=tg. CDE
b) when the anger ECD= angle DAB (interior angle)
angle BAD=angle ABD (opposite sides of triangle is equal)
In tg. BDC
angle DBC=angle DCB (opposite sides of triangle is equal)
so, tg. ABC =tg. ECB
c) we prove angle ABC = angle ECB is equal
when angle ABC is 90° so angle ECB is 90°
pls mark as brainly
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