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Answer 1

Step-by-step explanation:

Given information: AD is median, AB=AC, m∠B=35°.

In triangle ABC, two sides of triangle are equal. It means triangle ABC is an isosceles triangle.

$\angle B\cong \angle C∠B≅∠C (Isosceles \: triangle \: property) \\ </p><p></p><p>m\angle B=m\angle Cm∠B=m∠C (congruent \: angle \: property) \\ </p><p></p><p>35^{\circ}=m\angle C35 \degree\\ </p><p> =m∠C (Given, m∠B=35°)$

According to angle sum property, the sum of interior angle of a triangle is 180°.

In triangle ABC,

$\angle A+\angle B+\angle C=180^{\circ}∠A+∠B+∠C=180 \degree \\ </p><p> </p><p></p><p>\angle A+35^{\circ}+35^{\circ}=180^{\circ}∠A+35 \degree</p><p> +35 \degree</p><p> =180 \degree$

$\angle A+70^{\circ}=180^{\circ}∠A+70 </p><p>∘ </p><p> =180 \degree \\ </p><p> </p><p></p><p>\angle A=180^{\circ}-70^{\circ}∠A=180 \degree</p><p> −70 \degree \\ </p><p> </p><p></p><p>\angle A=110^{\circ}∠A=110 \degree</p><p>$

Median of an isosceles triangle divides the triangle in two equal parts.

In triangle ABC, AD is median. So,

$\triangle ABD\cong \triangle ACD△ABD≅△ACD \\ </p><p></p><p>\angle BAD\cong \angle CAD∠BAD≅∠CAD \\ </p><p></p><p>Angle \: A \: is \: the \: sum \: of \: angle \: BAD \: and \: CAD. \\ </p><p></p><p>\angle BAD+\angle CAD=\angle A∠BAD+∠CAD=∠A \\ </p><p></p><p>\angle BAD+\angle BAD=110^{\circ}∠BAD+∠BAD=110 \degree \\ </p><p> (\angle BAD\cong \angle CAD)(∠BAD≅∠CAD) \\ </p><p></p><p>2\angle BAD=110^{\circ}2∠BAD=110 </p><p>∘ \\ </p><p> </p><p></p><p>Divide \: both \: sides \: by 2.</p><p></p><p>\angle BAD=55^{\circ}∠BAD=55 \degree$