Math, asked by Anonymous, 6 months ago

1.
A ABC and A DBC are two isosceles triangles on
the same base BC and vertices A and D are on the
same side of BC (see Fig. 7.39). If AD is extended
to intersect BC at P, show that
(i) A ABD=AACD
(ii) A ABP=AACP
(iii) AP bisects Z A as well as D.
(iv) AP is the perpendicular bisector of BC.​

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Answered by mathdude500
1

Appropriate Question:

Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that

(i) Δ ABD ≅Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D.

(iv) AP is the perpendicular bisector of BC.

\large\underline{\sf{Solution-}}

Given that, Δ ABC and Δ DBC are two isosceles triangles on the same base BC.

\implies\sf \: \boxed{\sf \: AB = AC} \:  \: and \: \boxed{\sf \: DB = DC \: } \\  \\

Now, Consider

\sf \: In \:  \triangle\: ABD \: and \: \triangle\:ACD \\  \\

\boxed{\begin{aligned}&\sf \:AB=AC \:  \: (given)  \\ \\& \sf \:DB=DC  \:  \: (gien)\\ \\&\sf \: AD=AD \:  \: (common)\end{aligned}}\implies\sf \: \triangle\: ABD \cong\triangle\:ACD \\  \\

[ \sf \: By \: SSS \: Congruency \: rule \: ] \\  \\

\implies\sf \: \angle\:BAD = \angle\:CAD \:  \:  \: (CPCT) \\  \\

\implies\sf \: AD \: bisects \: \angle\:A \\  \\

\implies\sf \: \boxed{\sf \: AP \: bisects \: \angle\:A  \: }\\  \\

Now, Consider

\sf \: In \:  \triangle\: ABP \: and \: \triangle\:ACP\\  \\

\boxed{\begin{aligned}&\sf \:AB=AC \:  \: (given)  \\ \\& \sf \angle\:BAP=\angle\:CAP  \: (proved \: above)\\ \\&\sf \: AP=AP \:  \: (common)\end{aligned}}\implies\sf \: \triangle\: ABP \cong\triangle\:ACP\\  \\

[ \sf \: By \: SAS \: Congruency \: rule \: ] \\  \\

\implies\sf \:   \angle\: APB \:  =  \: \angle\:APC \: (CPCT) \: and \: BP = PC \: (CPCT)\\  \\

Now,

\implies\sf \:   \angle\: APB \:  +  \: \angle\:APC  =  {180}^{ \circ} \\  \\

\implies\sf \:   \angle\: APB \:  +  \: \angle\:APB  =  {180}^{ \circ} \\  \\

\implies\sf \: 2\angle\: APB  =  {180}^{ \circ} \\  \\

\implies\sf \: \angle\: APB  =  {90}^{ \circ} \\  \\

Thus, we concluded that

\sf \: AP \:  \perp \: BC \: and \: BP = PC \\  \\

\implies\sf \: \boxed{\sf \: AP \: is \: perperndicular \: bisector \: of \: BC \: } \\  \\

Now, Consider

\sf \: In \:  \triangle\: PDB \: and \: \triangle\:PDC \\  \\

\boxed{\begin{aligned}&\sf \:BD=DC \:  \: (given)  \\ \\& \sf PB = PC \: (proved \: above)\\ \\&\sf \: DP=DP \:  \: (common)\end{aligned}}\implies\sf \: \triangle\:PDB \cong\triangle\:PDC\\  \\

[ \sf \: By \: SSS \: Congruency \: rule \: ] \\  \\

\implies\sf \: \angle\:BDP = \angle\:CDP \:  \:  \: (CPCT) \\  \\

\implies\sf \: DP \: bisects \: \angle\:D \\  \\

\implies\sf \: \boxed{\sf \: AP \: bisects \: \angle\:D  \: }\\  \\

Answered by ayankaryan2007
1

Answer:

Please find the explanation in the attached photograph.

Step-by-step explanation:

Hope this helps you...

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