Question

Write a clear proof with statements and reasons

2. Prove theorem 1.7 (P. 25) for the following kite. Theorem 1 ne diagon al of a e vevticies wheve kile joining e 2 coanvet sides mee+ bisect ne ang les an neperpendicular bisector of e otner diogonal

Answer 1

Given ABCD be the kite with AB=BC and AD=CD

Join BD and AC to form diagonals

Extend BD to meet AC at M

To prove BD bisect $\angle B$ and BM bisect $\angle D$ and BM is the perpendicular bisector of AC at M

Consider $\bigtriangleup ADB and \bigtriangleup BDC$

$AB\cong BC$ Given AB=BC ,sides of kite

$BD\cong BD$ Common side

$AD\cong CD$ Given AD=CD ,sides of kite

$\bigtriangleup ADB\cong \bigtriangleup BDC$ bySSS axiom

$\angle ABD\cong \angle DBC$ Corresponding parts of congruent triangles are congruent

Hence BD bisects $\angle B$

ALso $\angle ADB\cong CDB$ Corresponding parts of congruent triangles..................................(1)

Consider $\bigtriangleup ADM and \bigtriangleup CDM$

$AD\cong CD$ given sides of kite

$\angle MAD\cong MCD$$\bigtriangleup ADC$ is isosceles and hence base angles are equal

$MA\cong MC$ the line joining vertex angle of an isosceles triangle to its base bisects the base

$\bigtriangleup ADM\cong \bigtriangleup CDM$ by SAS axiom

$\angle ADM\cong \angle CDM$ Corresponding parts of congruent triangles..................................(2)

From (1) and (2) BM bisects $\angle D$

Now to prove BM is the perpendicular bisector of AC

$\angle AMD\cong \angle CMD$ corresponding parts of congruent triangles

$\angle AMD+\angle CMD=180$ linear pair

$\angle AMD=\angle CMD=90$

Also AM=CM

hence BM is the perpendicular bisector of AC