In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

Question

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

KnowledgeBoat · Accepted Answer

(i) Since, D is mid-point of BC, ∴ CD = In right triangle ACD, By pythagoras theorem, ⇒ AD² = AC² + CD² ⇒ AD² = AC² + ⇒ AD² = AC² + ⇒ AD² = ⇒ 4AD² = 4AC² + BC² .....(1) Hence, proved that 4AD² = 4AC² + BC². (ii) As E is mid-point of AC, ∴ CE = ⇒ AC = 2CE BCE is right triangle, By pythagoras theorem, ⇒ BE² = BC² + CE² Multiplying both sides by 4 we get, ⇒ 4BE² = 4BC² + 4CE² ⇒ 4BE² = 4BC² + (2CE)² ⇒ 4BE² = 4BC² + AC² ......(2) Hence, proved that 4BE² = 4BC² + AC². (iii) As, ABC is a right triangle, By pythagoras theorem we get, ⇒ AB² = AC² + BC² Adding 1 and 2 from above parts we get, ⇒ 4AD² + 4BE² = 4AC² + BC² + 4BC² + AC² ⇒ 4(AD² + BE²) = 5AC² + 5BC² ⇒ 4(AD² + BE²) = 5(AC² + BC²) ⇒ 4(AD² + BE²) = 5AB². Hence, proved that 4(AD² + BE²) = 5AB².

Mathematics

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

Pythagoras Theorem

Related Questions

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the figure given below, ABCD is a square of side 7 cm. If
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

If AD, BE and CF are medians of △ABC, prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).

Mathematics

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :(i) 4AD2 = 4AC2 + BC2(ii) 4BE2 = 4BC2 + AC2(iii) 4(AD2 + BE2) = 5AB2

Pythagoras Theorem

Related Questions

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD2 = 3AB2.

In the figure given below, ABCD is a square of side 7 cm. IfAE = FC = CG = HA = 3 cm,(i) prove that EFGH is a rectangle.(ii) find the area and perimeter of EFGH.

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB2 + CD2 = AD2 + BC2.

If AD, BE and CF are medians of △ABC, prove that3(AB2 + BC2 + CA2) = 4(AD2 + BE2 + CF2).

In the adjoining figure, D and E are mid-points of the sides BC and CA respectively of a △ABC, right angled at C. Prove that :
(i) 4AD² = 4AC² + BC²
(ii) 4BE² = 4BC² + AC²
(iii) 4(AD² + BE²) = 5AB²

AD is perpendicular to the side BC of an equilateral △ABC. Prove that 4AD² = 3AB².

In the figure given below, ABCD is a square of side 7 cm. If
AE = FC = CG = HA = 3 cm,
(i) prove that EFGH is a rectangle.
(ii) find the area and perimeter of EFGH.

In the adjoining figure, the diagonals AC and BD of a quadrilateral ABCD intersect at O, at right angles. Prove that AB² + CD² = AD² + BC².

If AD, BE and CF are medians of △ABC, prove that
3(AB² + BC² + CA²) = 4(AD² + BE² + CF²).