Question

​​If D is the midpoint of the hypotenuse AC of a right triangle ABC, prove that BD = 1/2AC

Answer 1

please mark as BRAINLIEST

Answer 2

Given:

• $D$ is the midpoint of the hypotenuse $AC$.
• Δ$ABC$ is a right angled triangle, right angled at $B$.

To find:

• Is $BC=\frac{1}{2}AC$.

Step-by-step explanation:

• First of all we need to construct and produce $BD$ to $E$ such that,

        $BD=DE$

• Now join $EC$.
• consider two triangles Δ$ADB$ and Δ$CDE$.
• In Δ$ADB$ and Δ$CDE$,
• It is given that,

        $AD=CD$       -----------$(1)$

• By construction,

        $BD=DE$      ------------$(2)$

• We know that vertically opposite angles are equal so,

        ∠$ADB=$∠$CDE$   ---------$(3)$

• By SAS that is Side angle Side criterion of congruence

        Δ$ADB$≅Δ$CDE$

• Therefore by CPCT,

        $EC=AB$    -----------$(A)$

• And,

        ∠$CED=$∠$ABD$  --------------$(A.1)$

• $CED$ and $ABD$ are alternate interior angles.
• Since alternate interior angles are equal, so,

        $CE$║$AB$

• We know that sum of the alternate interior angle is $180$° so,

        ∠$ABC+$∠$ECB=180$°

• We know that Δ$ABC$ is right angle.

        $90$°$+$∠$ECB=180$°

• By solving the above equation we get,

        ∠$ECB=90$°

• In the triangles Δ$ABC$ and Δ$ECB$,
• We know that,

        $AB=EC$

        $BC=CB$ (common)

• The right angles are equal,

        ∠$ABC=$∠$ECB=90$°

• So by Side angle side (SAS) criterion of congruence,

        Δ$ABC$≅Δ$ECB$

• Therefore by CPCT

        $AC=BE$

        $\frac{1}{2}AC=\frac{1}{2}BE$

• Thus,

        $BD=\frac{1}{2}AC$

Final answer:

• Thus $BD=\frac{1}{2}AC$ has been proved.