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Question
- given: mt and ha intersect at b, ma||ht, and mt bisects ha. prove: ma≅ht
Step1: Use the property of parallel - lines and angle - bisector
Since $\overline{MA}\parallel\overline{HT}$ and $\overline{MT}$ bisects $\overline{HA}$ at point $B$, we know that $\angle MBA=\angle BHT$ (alternate - interior angles) and $\angle AMB=\angle MTH$ (alternate - interior angles). Also, because $\overline{MT}$ bisects $\overline{HA}$, we have $HB = BA$.
Step2: Prove triangle congruence
In $\triangle MAB$ and $\triangle THB$, we have:
- $\angle MBA=\angle BHT$ (alternate - interior angles from $\overline{MA}\parallel\overline{HT}$)
- $HB = BA$ (given that $\overline{MT}$ bisects $\overline{HA}$)
- $\angle AMB=\angle MTH$ (alternate - interior angles from $\overline{MA}\parallel\overline{HT}$)
By the Angle - Side - Angle (ASA) congruence criterion, $\triangle MAB\cong\triangle THB$.
Step3: Use the property of congruent triangles
Since $\triangle MAB\cong\triangle THB$, the corresponding sides are equal. So, $\overline{MA}\cong\overline{HT}$.
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$\overline{MA}\cong\overline{HT}$ is proved.