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Prove That: (I) ∆ Abd ≅ ∆ Acd (Ii) ∠B = ∠C (Iii) ∠Adb = ∠Adc (Iv) ∠Adb = 90° - Mathematics

Sum

Prove that:
(i) ∆ ABD ≅ ∆ ACD
(ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC
(iv) ∠ADB = 90°

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Solution

Given: In the figure,
AD = AC 
BD = CD

To prove:
(i) Δ ABD ≅ Δ ACD
(ii) ∠B = ∠C
(iii) ∠ADB = ∠ADC
(iv) ∠ADB = 90°

Proof: In Δ ABD and Δ ACD
AD = AD .........(common)
AD = AC ..........(given)
BD = CD ...........(given)

(i) ∴ Δ ABD ≅ Δ ACD ........(SSS axiom)

(ii) ∴ ∠B = ∠C .............(c.p.c.t.)

(iii) ∠ADB = ∠ADC ..........(c.p.c.t.)

But ∠ADB + ∠ADC = 180° ..........(Linear pair)

∴ ∠ADB = ∠ADC

(iv) ∠ADB = ∠ADC

= `(180°)/2`

= 90°

  Is there an error in this question or solution?
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APPEARS IN

Selina Concise Mathematics Class 7 ICSE
Chapter 19 Congruency: Congruent Triangles
Exercise 19 | Q 4
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