Question

Draw a right triangle and the perpendicular from the midpoint of the hypotenuse to the base.

i. Prove that this perpendicular is half the perpendicular side of the large triangle.

ii. Prove that perpendicular bisects the bottom side of the larger triangle.

iii. Prove that in the large triangle the distances from the mid point of the hypotenuse to all the vertices are equal.

iv. Prove that the circumcentre of a right triangle is the mid point of its hypotenuse.

Answer 1

MN is perpendicular to AB.

MN and CB are parallel lines. MN divides AC and AB in the same ratio.

i. AM = \(\frac{1}{2}\)AC, AN = \(\frac{1}{2}\) AB

∴ MN = \(\frac{1}{2}\)CB

ii. We get two right triangles of same base and perpendicular. So their hypotenuses are also equal.

∴ MA = MC = MB

AN = NB

iii. ∠ANM = ∠BNM = 90°

MN = MN

ΔANM ≅ ΔBNM

AM = MB …….(1)

M is the midpoint of AC

AM = MC …….(2)

(1) = (2) ⇒ AM = MB = MC

iv. The points A, B and C are at same distance from M. So a circle can be drawn with M as centre and pass through there three points. So this circle is the circum circle of ΔABC.