Question

P is the midpoint of hypothenuse AB in a right angled triangle ABC prove AB= 2CP find using midpoint theorems​

Answer 1

✪ᴇxᴘʟᴀɪɴᴀᴛɪᴏɴ✪

☆ɢɪᴠᴇɴ☆

• Right triangle ABC, right angled at C with P as the mid point of hypotenous AB.

☆ᴛᴏ ᴘʀᴏᴠᴇ☆

• $AB = 2CP$

☆ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ☆

• Draw PD and PE parallel to BC and AC respectively.

☆ᴘʀᴏᴏғ☆

Mid Point Theorem

If a line is drawn from mid point of one side of a triangle, parallel to another side and intersects the third side then it will bisect the third side of the triangle.

By mid point theorem

$AD = DC, BE = EC .....(1)$

Also, $PD||BC, PE||AC$ but $AC⊥BC$,

$⇒PD ⊥ AC, PE ⊥BC$

$⇒ PDCE$ is a rectangle

$⇒ DC = PE .....(2)$

Again PE⊥BC, so ΔPCE is right triangle. Applying Pythagoras Theorem,

$CP² = EC² + PE² .....(3)$

Applying Pythagoras Theorem in ΔABC,

$AB² = AC² + BC²$

$⇒AB² = (AD+DC)² + (BE+EC)²$

$⇒AB² = (2DC)² + (2EC)²$[Using (1)]

$⇒AB² = 4(DC)² + 4(EC)²$

$⇒AB² = 4(PE)² + 4(EC)²$[Using (2)]

$⇒AB² = 4[(PE)² + 4(EC)²]$

$⇒AB² = 4(CP)²$ [Using (3)]

$⇒AB = 2CP$

Hence Proved