### Prove two conquent triangles in a square; prove one triangle is right-angled and then find its length

Question Sample Titled 'Prove two conquent triangles in a square; prove one triangle is right-angled and then find its length'

In the figure, ${A}{B}{C}{D}$ is a square. ${E}$ and ${F}$ are points lying on ${B}{C}$ and ${C}{D}$ respectively such that ${A}{E}={B}{F}$. ${A}{E}$ and ${B}{F}$ intersect at ${G}$.

${G}$${A}$${B}$${C}$${D}$${E}$${F}$

 (a) Prove that $\triangle{A}{B}{E}\stackrel{\sim}{=}\triangle{B}{C}{F}$ . (2 marks) (b) Is $\triangle{B}{G}{E}$ a right-angled triangle? Explain your answer. (3 marks) (c) If ${C}{F}={10}$ $\text{cm}$ and ${E}{G}={6}$ $\text{cm}$ , find ${B}{G}$ . (2 marks)

 (a)  ${A}{B}$ $={B}{C}$ property of square ${A}{E}$ $={B}{F}$ given $\angle{A}{B}{E}$ $={90}^{\circ}=\angle{B}{C}{F}$ property of square $\triangle{A}{B}{E}$ $\stackrel{\sim}{=}\triangle{B}{C}{F}$ RHS  Marking scheme of (a): Case 1 Any correct proof with correct reasons. 2M Case 2 Any correct proof without reasons. 1M  (b) Let $\angle{B}{A}{E}$ $={x}$ . $\angle{C}{B}{F}$ $=\angle{B}{A}{E}={x}$ corr. ∠s ≅△s 1M ∴  $\angle{A}{B}{G}$ $={90}^{\circ}-{x}$ $\angle{B}{G}{E}$ $={x}+{\left({90}^{\circ}-{x}\right)}$ ext. ∠ of △ 1M  $={90}^{\circ}$ Thus, $\triangle{B}{G}{E}$ is a right-angled triangle. 1A f.t.  (c) ${B}{E}={C}{F}={10}$ $\text{cm}$ corr. sides, ≅△s ${B}{G}$ $=\sqrt{{{B}{E}^{{2}}-{E}{G}^{{2}}}}$ 1M $=\sqrt{{{10}^{{2}}-{6}^{{2}}}}$ $={8}$ $\text{cm}$ 1A

${G}$${A}$${B}$${C}$${D}$${E}$${F}$${x}$${x}$${90}^{\circ}-{x}$

 (b) Let $\angle{B}{E}{A}={x}$ . $\angle{C}{F}{B}$ $=\angle{B}{E}{A}={x}$ corr. ∠s ≅△s 1M ∴  ${G}$ , ${E}$ , ${C}$ and ${F}$ are concyclic. ext. ∠ = int. opp. ∠ 1M $\angle{B}{G}{E}$ $=\angle{E}{C}{F}={90}^{\circ}$ ext. ∠, cyclic quad. Thus, $\triangle{B}{G}{E}$ is a right-angled triangle. 1A f.t.

${G}$${A}$${B}$${C}$${D}$${E}$${F}$${x}$${x}$

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