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prove that ∠abc is a right - angle. given: ad≅bd≅cd. prove: m∠abc = 90.…

Question

prove that ∠abc is a right - angle. given: ad≅bd≅cd. prove: m∠abc = 90. complete the proof below. statement: ad≅bd≅cd, △adb and △cdb are isosceles. reason: given

Explanation:

Step1: Use isosceles - triangle property

In $\triangle ADB$, since $AD\cong BD$, let $\angle A=\angle ABD = x$. In $\triangle CDB$, since $BD\cong CD$, let $\angle C=\angle CBD = y$.

Step2: Apply angle - sum property of a triangle

In $\triangle ABC$, the sum of interior angles is $180^{\circ}$. So $m\angle A + m\angle ABC+m\angle C=180^{\circ}$. And $m\angle ABC=x + y$.

Step3: Substitute and solve

Substituting $\angle A=x$, $\angle ABC=x + y$, and $\angle C=y$ into the angle - sum formula, we get $x+(x + y)+y = 180^{\circ}$, which simplifies to $2(x + y)=180^{\circ}$, then $x + y = 90^{\circ}$. Since $m\angle ABC=x + y$, $m\angle ABC = 90^{\circ}$.

Answer:

$m\angle ABC = 90^{\circ}$ is proved as above.