1. **Problem Statement:** Given isosceles triangle $PQR$ with $PQ = PR$, points $S$ and $T$ lie on $PQ$ and $PR$ respectively such that $ST \parallel QR$. Perpendiculars from $S$ and $T$ to $QR$ meet at $A$ and $B$ respectively. Also, $RS$ is perpendicular to $PQ$. We need to prove: ii. $PT = PS$ iii. $\triangle QSA = \triangle TRB$ iv. $\triangle QSA$ and $\triangle QSR$ are equiangular v. $QS^2 = QR \cdot AQ$ 2. **Key Properties and Formulas:** - Since $PQ = PR$, $\triangle PQR$ is isosceles. - $ST \parallel QR$ implies $\triangle PST \sim \triangle PQR$ by AA similarity. - Perpendiculars imply right angles at $A$ and $B$. - Equiangular triangles have equal corresponding angles. 3. **Proof of (ii) $PT = PS$: ** - From similarity $\triangle PST \sim \triangle PQR$, corresponding sides are proportional: $$\frac{PS}{PQ} = \frac{PT}{PR}$$ - Since $PQ = PR$, this reduces to: $$PS = PT$$ 4. **Proof of (iii) $\triangle QSA = \triangle TRB$: ** - $A$ and $B$ are feet of perpendiculars from $S$ and $T$ to $QR$, so $\angle QAS = \angle TRB = 90^\circ$. - $ST \parallel QR$ implies alternate interior angles $\angle QSA = \angle TRB$. - $QS = TR$ by similarity and isosceles property. - By AA similarity and side equality, $\triangle QSA \cong \triangle TRB$. 5. **Proof of (iv) $\triangle QSA$ and $\triangle QSR$ are equiangular: ** - $RS$ is perpendicular to $PQ$, so $\angle QSR = 90^\circ$. - $\angle QSA$ is also right angle (perpendicular from $S$ to $QR$). - $\angle QAS = \angle QRS$ (alternate interior angles due to parallel lines). - Hence, all corresponding angles equal, so triangles are equiangular. 6. **Proof of (v) $QS^2 = QR \cdot AQ$: ** - In right triangle $QSA$, by similarity with $QRA$ (right triangle with altitude $AQ$), the geometric mean property holds: $$QS^2 = QR \times AQ$$ **Final answers:** ii. $PT = PS$ iii. $\triangle QSA = \triangle TRB$ iv. $\triangle QSA$ and $\triangle QSR$ are equiangular v. $QS^2 = QR \cdot AQ$