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Question
In ΔABC, side AB is produced to D so that BD = BC. If ∠B = 60° and ∠A = 70°, prove that: (i) AD > CD (ii) AD > AC
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Solution
Given that in ΔABC,side AB is produced to D So that BD = BC and ∠B =60°,∠A= 70°
We have to prove that
(1)AD>CD (2)AD>AC
First join C and D
Now, in ΔABC
∠A+∠B+∠C=180° [ ∵ Sum of angles in a triangle 180° ]
⇒ ∠C=180°-70°-60°
=180°-130°=50°
∴ ∠C=50° ⇒ ∠ACB=50° ..............(1)
And also in ,Δ BDC,
∠DBC=180°-∠ABC [∵ABDis a straight angle ]
=180°-60°=120°
and also BD=BC [given]
⇒ ∠BCD=∠BDC [ ∵Angles opposite to equal sides are equal]
Now,
∠DBC+∠BCD+∠BDC=180° [ ∵Sum of angles in a triangle 180°]
⇒ 120°+∠BCD+∠BCD=180°
⇒2∠BCD=180°-120°⇒ ∠BCD=`(60°) /2=30`
∴ ∠BCD=∠BDC=30° ..............(2)
Now, consider , ΔADC,
∠BAC⇒∠DAC=70 [given]
∠BDC⇒∠ADC=30 [ ∵ From (2)]
∠ACD=∠ACB+∠BCD
=50°+ 30° [ ∵ From (1) and (2)]
=80°
Now, ∠ADC<∠DAC<ACD
⇒ AC<DC<AD
[∵ Side opposite to greater angle is longer and smaller angle is smaller]
⇒ AD>DC and AD>AC
∴Hence proved
Or
We have, ∠ACD>∠DAC and ∠ACD>∠ADC
⇒ AD>DC and AD>AC
[ ∵ Side opposite to greater angle is longer and smaller angle is smaller]
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