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Question
given: △abc; $overline{ca}congoverline{cb}$; $overline{ad}$ and $overline{be}$ are altitudes of △abc. prove: $overline{ad}congoverline{be}$ 1. △abc; $overline{ca}congoverline{cb}$; $overline{ad}$ and $overline{be}$ are altitudes of △abc. 1. given 2. definition of an altitude 3. all right angles are congruent. 4. ∠abc≅∠bac
Step1: State given information
Given $\triangle ABC$ with $\overline{CA}\cong\overline{CB}$ and $\overline{AD},\overline{BE}$ as altitudes.
Step2: Define right - angles
Since $\overline{AD}$ and $\overline{BE}$ are altitudes, $\angle ADB = \angle BEC = 90^{\circ}$.
Step3: Use right - angle congruence
All right - angles are congruent, so $\angle ADB\cong\angle BEC$.
Step4: Identify base - angles of isosceles triangle
In $\triangle ABC$, because $\overline{CA}\cong\overline{CB}$, $\angle ABC\cong\angle BAC$ (base - angles of an isosceles triangle).
Step5: Apply reflexive property
$\overline{AB}$ is congruent to itself, i.e., $\overline{AB}\cong\overline{BA}$.
Step6: Prove triangle congruence
By AAS congruence criterion (two angles and a non - included side), $\triangle ADB\cong\triangle BEC$.
Step7: Use CPCTC
Since the triangles are congruent, corresponding parts are congruent, so $\overline{AD}\cong\overline{BE}$.
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- $\triangle ABC; \overline{CA}\cong\overline{CB}; \overline{AD}$ and $\overline{BE}$ are altitudes of $\triangle ABC$. - Given
- $\angle ADB = \angle BEC=90^{\circ}$ - Definition of an altitude
- $\angle ADB\cong\angle BEC$ - All right - angles are congruent
- $\angle ABC\cong\angle BAC$ - Base - angles of an isosceles triangle are congruent ($\triangle ABC$ is isosceles since $\overline{CA}\cong\overline{CB}$)
- $\overline{AB}\cong\overline{BA}$ - Reflexive property of congruence
- $\triangle ADB\cong\triangle BEC$ - AAS (Angle - Angle - Side) congruence criterion ($\angle ADB\cong\angle BEC$, $\angle BAD\cong\angle ABE$, $\overline{AB}\cong\overline{BA}$)
- $\overline{AD}\cong\overline{BE}$ - Corresponding parts of congruent triangles are congruent (CPCTC)