In the given triangle, ∠AED ∼ ∠ ABC, AD = 6.9, AE = 7.2, DE = 5.2, and BC = 10.2. Find the measure of BD and CE. Round your answer to the nearest tenth.

Answer:The measure of side BD is 8.6  and The measure of side CE is 8.4  Step-by-step explanation:Given as :The Triangle is ABC with side AB , BC , CAAnd The points E  and D  is on the side AB and ACSo, AED is a TriangleAnd Δ AED [tex]\sim[/tex] Δ ABCThe measure of side AD = 6.9The measure of side AE = 7.2The measure of side ED = 5.2The measure of side BC = 10.2Let The The measure of side EB = xAnd The measure of side DC = ySo, From similarity property[tex]\dfrac{AB}{AE}[/tex] = [tex]\dfrac{AC}{AD}[/tex] = [tex]\dfrac{BC}{ED}[/tex]Or,  [tex]\dfrac{AB}{AE}[/tex] = [tex]\dfrac{BC}{ED}[/tex]So, [tex]\dfrac{7.2 + x}{7.2}[/tex] = [tex]\dfrac{10.2}{5.2}[/tex]Or, 5.2 × ( 7.2 + x ) = 10.2 × 7.2Or, 37.44 + 5.2 x = 73.44Or, 73.44 - 37.44 = 5.2 x∴ x = [tex]\frac{36}{5.2}[/tex]I.e x = 6.9 Now in Δ BEDBE² + ED² = BD²Or, 6.9² + 5.2² = BD²Or, BD² = 74.65∴  BD = [tex]\sqrt{74.65}[/tex]I.e BD = 8.64Or, BD = 8.6Similarly for y [tex]\dfrac{AC}{AD}[/tex] = [tex]\dfrac{BC}{ED}[/tex]Or,  [tex]\dfrac{6.9+y}{6.9}[/tex] = [tex]\dfrac{10.2}{5.2}[/tex]Or, 5.2 × ( 6.9 + y ) = 10.2 × 6.9Or, 35.88 + 5.2 y = 70.38or, 5.2 y = 70.8 - 35.88Or, 5.2 y = 34.5∴  y = [tex]\frac{34.5}{5.2}[/tex]I.e y = 6.6Now in Δ CEDCD² + ED² = CE²Or, 6.6² + 5.2² = CE²Or, CE² = 70.6∴  CE = [tex]\sqrt{70.6}[/tex]I.e CE = 8.40Or, CE = 8.4Hence The measure of side BD is 8.6  and The measure of side CE is 8.4  Answer