In ∆ABC, ∠BAC = 90°, seg BL and seg CM are medians of ∆ABC. Then prove that: 4(BL2 + CM2) = 5 BC2

solution Given: $$\Delta ABC$$ right angled at $$A$$ i.e; $$A = 90^\circ$$. where $$BL$$ and $$CM$$ are the median. To Prove: $$4( BL^2 + CM^2 ) = 5BC^2$$ Proof: Since $$BL$$ is the median, $$AL = CL = \frac{1}{2} AC \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1)$$ Similarly, $$CM$$ is the median $$AM = MB = \frac{1}{2}AB \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(2)$$ We know that, by Pythagoras' theorem $$(\text{Hypotenuse})^2 = (\text{Height})^2 + (\text{Base})^2 \qquad\qquad(3)$$ In $$\Delta BAC$$, $$(BC)^2 = (AB)^2 + (AC)^2\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(4)$$ In $$\Delta BAL$$, $$(BL)^2 = AB^2 + AL^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(\text{From }1)$$ $$\qquad\qquad\quad= AB^2 + (\frac{AC}{2})^2$$ $$\qquad\qquad\quad= AB^2 + \frac{AC^2}{4}$$ $$\qquad\qquad\quad= \frac{4AB^2 + AC^2}{4}$$ $$\therefore\quad 4BL^2 = 4AB^2 + AC^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(5)$$ In $$\Delta MAC$$, $$(CM)^2 = (AM)^2 + (AC)^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad(\text{From }2)$$ $$\qquad\qquad\quad= (\frac{AB}{2})^2 + (AC)^2$$ $$\qquad\qquad\quad= \frac{AB^2 + 4AC^2}{4}$$ $$\therefore\quad 4CM^2 = AB^2 + 4AC^2 \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(6)$$ From $$(4), (5)$$ and $$(6)$$, $$(BC)^2 = (AB)^2 + (AC)^2$$ $$4(BL)^2 = 4(AB)^2 + (AC)^2$$ $$4(CM)^2 = (AB)^2 + 4(AC)^2$$ Adding $$(5)$$ and $$(6)$$, $$4(BL)^2 + 4(CM)^2 = 4(AB)^2 + (AC)^2 + (AB)^2 + 4(AC)^2$$ $$4(BL^2 + CM^2) = 5(AB^2 + AC^2)$$ $$4(BL^2 + CM^2) = 5(BC)^2$$ Hence Proved.