Question
In the given figure; ∠1=∠2 and AB = AC. Prove that:  [4 MARKS]
(i) ∠B=∠C
(ii) BD = DC
(iii) AD is perpendicular to BC.
(i) ∠B=∠C
(ii) BD = DC
(iii) AD is perpendicular to BC.
Answer:
:
Steps: 1 Mark
Each proof: 1 Mark
In ΔADB and ΔADC
∠1=∠2 [Given]
⇒∠BAD=∠CAD
AD=AD [Common side]
AB=AC [Given]
⇒ΔADB≅ΔADC [SAS congruency criteria]
(i) ∴∠B=∠C [Corresponding parts of congruent triangles]
(ii) BD=DC [Corresponding parts of congruent triangles]
(iii)ΔADB≅ΔADC [proved above]
∠ADB+∠ADC=180° [Linear pair]
⇒∠ADB=∠ADC [c.p.c.t]
∴∠ADB+∠ADB=180°
⇒2∠ADB=180°
⇒∠ADB=180°2
=90°
⇒AD⊥BC
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:
Steps: 1 Mark
Each proof: 1 Mark
In ΔADB and ΔADC
∠1=∠2 [Given]
⇒∠BAD=∠CAD
AD=AD [Common side]
AB=AC [Given]
⇒ΔADB≅ΔADC [SAS congruency criteria]
(i) ∴∠B=∠C [Corresponding parts of congruent triangles]
(ii) BD=DC [Corresponding parts of congruent triangles]
(iii)ΔADB≅ΔADC [proved above]
∠ADB+∠ADC=180° [Linear pair]
⇒∠ADB=∠ADC [c.p.c.t]
∴∠ADB+∠ADB=180°
⇒2∠ADB=180°
⇒∠ADB=180°2
=90°
⇒AD⊥BC
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