In the figure given below- - DAB-100- and - DBC-30- Find- BDC-

Consider ADB and BDC, BD = BD..... (common side) AD = DC..... nbsp; (given) BA = BC..... nbsp; (given) ⇒ ADB ≅ BDC (By SSS criteria) Consider DBC, ⇒∠ DCB ...

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Byju's Answer

Standard VIII

Mathematics

Medians of a Triangle and the Drawing Method

In the figure...

Question

In the figure given below, $∠ D A B = 100^{\circ} a n d ∠ D B C = 30^{\circ}$ . Find $∠ B D C$ .

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Solution

Consider $△ A D B a n d △ B D C,$
BD = BD..... (common side)
AD = DC..... (given)
BA = BC..... (given)
$\Rightarrow △ A D B$ $≅$ $△ B D C$ (By SSS criteria)

Consider $△ D B C,$
$\Rightarrow ∠ D C B = ∠ D A B = 100^{\circ}$ ....(congruent parts of congruent triangle)
$∠ D B C = 30 °$ ....... (given)

$∠ D C B + ∠ C B D + ∠ B D C = 180 °$ ....(Angle sum property of triangle)
$\Rightarrow 100 ° + 30 ° + ∠ B D C = 180 °$
$\Rightarrow 130 ° + ∠ B D C = 180 °$
$\Rightarrow$ $∠ B D C = 180 ° - 130 °$
$\Rightarrow$ $∠ B D C = 50 °$

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