6. In Fig. 6.37 , if \( \triangle A B E \cong \triangle A C D \), show that \( \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC} \). 7. In Fig. 6.38, altitudes AD and CE of \( \triangle \mathrm{ABC} \) intersect each other at the point \( P \). Show that: (i) \( \triangle \mathrm{AEP} \sim \triangle \mathrm{CDP} \) (ii) \( \triangle \mathrm{ABD} \sim \triangle \mathrm{CBE} \) (iii) \( \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB} \) (iv) \( \triangle \mathrm{PDC} \sim \triangle \mathrm{BEC} \) Fig. 6.37
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Let's address each part step by step. **6. Show that \( \triangle \mathrm{ADE} \sim \triangle \mathrm{ABC} \):** Given that \( \triangle ABE \cong \triangle ACD \), we know that all corresponding sides and angles of these triangles are equal. Therefore, we Show more…
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Key Concepts
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In Fig. $6.38$, altitudes AD and $\mathrm{CE}$ of $\triangle \mathrm{ABC}$ intersect each other at the point $\mathrm{P}$. Show that: (i) $\triangle \mathrm{AEP}-\Delta \mathrm{CDP}$ (ii) $\Delta \mathrm{ABD}-\Delta \mathrm{CBE}$ (iii) $\Delta \mathrm{AEP}-\Delta \mathrm{ADB}$ (iv) $\Delta \mathrm{PDC}-\Delta \mathrm{BEC}$
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In Fig. $6.39, \mathrm{ABC}$ and $\mathrm{AMP}$ are two right triangles, right angled at $\mathrm{B}$ and $\mathrm{M}$ respectively. Prove that: (i) $\Delta \mathrm{ABC}-\Delta \mathrm{AMP}$ (ii) $\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}$
In Fig. $6.60, \mathrm{AD}$ is a median of a triangle $\mathrm{ABC}$ and $\mathrm{A} M \perp \mathrm{BC}$. Prove that : (i) $\mathrm{AC}^{2}=\mathrm{AD}^{2}+\mathrm{BC} \cdot \mathrm{DM}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$ (ii) $\mathrm{AB}^{2}=\mathrm{AD}^{2}-\mathrm{BC} \cdot \mathrm{DM}+\left(\frac{\mathrm{BC}}{2}\right)^{2}$ (iii) $\mathrm{AC}^{2}+\mathrm{AB}^{2}=2 \mathrm{AD}^{2}+\frac{1}{2} \mathrm{BC}^{2}$
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