Question

Please prove?

AB is the diameter of the circle.
DE is perpendicular on AB.
Arc AC=arc DB
To prove angle ABC=angle BDE

Answer 1

Given: AB is the diameter, arc AC= arcDB, `DE_|_AB`

RTP: `angleABC=angleBDE`

Construction: . `A,D` are joined.

Now in `DeltaABD and DeltaBED`

`angle ADB=angle BED=90^@` as `angle ADB` is semicircular and `DE_|_AB` is drawn. `angle DBE=angle ABD` common

Hence remaining `color(red)(angle BAD=angle BDE......[1])`

So `DeltaABD and DeltaBED` are SIMILAR

Again in `DeltaABC and DeltaABD`

`AC=BD` as it is given arc AC= arc DB

`angleACB and angleABD =90^ @` as they are semicircular angles.
and `AB` hypotenuse is common.

So `DeltaABC~= DeltaABD`

Hence `color(red)(angle BAD=angle ABC......[2])`

Combining [1] and [2] we get `color(magenta)(angleABC=angleBDE)`