In the figure below, triangle $ABC$ is a right-angled isosceles triangle, where \(\angle ABC=

Question

In the figure below, triangle $ABC$ is a right-angled isosceles triangle, where $\angle ABC={{90}^{\circ }}$ and $AB=BC$. Given that $D$ is a point on $AC$, such that $BD=CD$. Show that triangle $ABD$ is also a right-angled isosceles triangle. Show that $D$ is the mid-point of $AC$. Hence, state the relationship between the lengths of $BD$ and $AC$. In the figure below, triangle $ABC$ is a right-angled triangle, where $\angle ABC={{90}^{\circ }}$ and $\angle ACB={{30}^{\circ }}$. Given that $D$ is a point on $AC$, such that $BD=CD$. Show that triangle $ABD$ is an equilateral triangle. Show that $D$ is the mid-point of $AC$. Hence, state the relationship between the lengths of $BD$ and $AC$. In the figure below, triangle $ABC$ is a right-angled triangle, where $\angle ABC={{90}^{\circ }}$. Given that $D$ is a point on $AC$, such that $BD=CD$. Determine the type of triangle $ABD$, stating your reason clearly. Show that $D$ is the mid-point of $AC$. Hence, state the relationship between the lengths of $BD$ and $AC$.

Accepted Answer

(i)(a) Triangle $ABD$ is right-angled isosceles (i)(b) $D$ is the midpoint of $AC$ (i)(c) $BD=\frac12 AC$ (ii)(a) Triangle $ABD$ is equilateral (ii)(b) $D$ is the midpoint of $AC$ (ii)(c) $BD=\frac12 AC$ (iii)(a) Triangle $ABD$ is isosceles with $AD=BD$ (iii)(b) $D$ is the midpoint of $AC$ (iii)(c) $BD=\frac12 AC$

IP1U12 HOT Q1

IP1U12 HOT Q1

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